Question: $\lim_{x\to\infty}\dfrac{e^{x^2}}{x^4}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $\dfrac{1}{4}$ (Choice C) C $\dfrac{1}{2}$ (Choice D) D $\infty$
Solution: $\lim_{x\to\infty} e^{x^2}=\infty$ and $\lim_{x\to\infty} x^4=\infty$, so $\lim_{x\to\infty}\dfrac{e^{x^2}}{x^4}$ results in the indeterminate form $\dfrac{\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{e^{x^2}}{x^4} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[e^{x^2}\right]}{\dfrac{d}{dx}[x^4]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{2x\cdot e^{x^2}}{4x^3} \\\\ &=\lim_{x\to\infty}\dfrac{e^{x^2}}{2x^2} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[e^{x^2}\right]}{\dfrac{d}{dx}[2x^2]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{2x\cdot e^{x^2}}{4x} \\\\ &=\lim_{x\to\infty}\dfrac{e^{x^2}}{2} \\\\ &=\infty \end{aligned}$ Note that we used l'Hôpital's rule twice, because the first time we used it, we ended with the indeterminate form $\dfrac{\infty}{\infty}$ too. Also note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[e^{x^2}\right]}{\dfrac{d}{dx}[2x^2]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{e^{x^2}}{x^4}=\infty$.